Chapter 11: Conic Sections Exercise 11.1

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through points $(2,3)$ and $(-1,1)$,

${{(2-h)}^{2}}+{{(3-k)}^{2}}={{r}^{2}}$ ………(i)

${{(-1-h)}^{2}}+{{(1-k)}^{2}}={{r}^{2}}$ ………(ii)

Since the centre (h,k) of the circle lies on line $x-3y-11=0$

$h-3k=11$ ………(iii)

From equations (i) and (ii), we get

$\Rightarrow {{(2-h)}^{2}}+{{(3-k)}^{2}}={{(-1-h)}^{2}}+{{(1-k)}^{2}}$

$\Rightarrow 4-4h+{{h}^{2}}+9-6k+{{k}^{2}}=1+2h+{{h}^{2}}+1-2k+{{k}^{2}}$

$\Rightarrow 4-4h+9-6k=1+2h+1-2k$

$\Rightarrow 6h+4k=11$ ………(iv)

On solving equations (iii) and (iv), we obtain $h=\dfrac{7}{2}$ and $k=\dfrac{-5}{2}$

On substituting the values of h and k in equation (i), we obtain

$\Rightarrow {{\left( 2-\dfrac{7}{2} \right)}^{2}}+{{\left( 3+\dfrac{5}{2} \right)}^{2}}={{r}^{2}}$

$\Rightarrow {{\left( \dfrac{4-7}{2} \right)}^{2}}+{{\left( \dfrac{6+5}{2} \right)}^{2}}={{r}^{2}}$

$\Rightarrow {{\left( \dfrac{-3}{2} \right)}^{2}}+{{\left( \dfrac{11}{2} \right)}^{2}}={{r}^{2}}$

$\begin{align}   & \Rightarrow \dfrac{9}{4}+\dfrac{121}{4}={{r}^{2}} \\  & \Rightarrow \dfrac{130}{4}={{r}^{2}} \\ \end{align}$

Thus, the equation of the required circle is

${{\left( x-\dfrac{7}{2} \right)}^{2}}+{{\left( y+\dfrac{5}{2} \right)}^{2}}=\dfrac{130}{4}$

$\begin{align}   & {{\left( \dfrac{2x-7}{2} \right)}^{2}}+{{\left( \dfrac{2y+5}{2} \right)}^{2}}=\dfrac{130}{4} \\ & 4{{x}^{2}}-28x+49+4{{y}^{2}}+20y+25=130 \\ \end{align}$

$\begin{align}   & 4{{x}^{2}}+4{{y}^{2}}-28x+20y-56=0 \\  & 4({{x}^{2}}+y{}^{2}-7x+5y-14)=0 \\ \end{align}$

${{x}^{2}}+{{y}^{2}}-7x+5y-14=0$

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the radius of the circle  is 5 and its centre lies on the x-axis, k=0 and r=5.

Now, the equation of the circle passes through point $(2,3).$

$\begin{align}   & \therefore {{\left( 2-h \right)}^{2}}+{{3}^{2}}=25 \\  & \Rightarrow {{\left( 2-h \right)}^{2}}=25-9 \\  & \Rightarrow {{\left( 2-h \right)}^{2}}=16 \\ \end{align}$

$\begin{align}  & \Rightarrow \left( 2-h \right)=\pm \sqrt{16} \\  & =\pm 4 \\ \end{align}$

If $2-h=4,$ then $h=-2$

If $2-h=-4,$ then $h=6$

When $h=-2$, the equation of the circle becomes

$\begin{align}  & {{\left( x+2 \right)}^{2}}+{{y}^{2}}=25 \\  & {{x}^{2}}+4x+4+{{y}^{2}}=25 \\  & {{x}^{2}}+{{y}^{2}}+4x-21=0 \\ \end{align}$

When $h=6$, the equation of the circle becomes

$\begin{align}   & {{\left( x-6 \right)}^{2}}+{{y}^{2}}=25 \\  & {{x}^{2}}-12x+36+{{y}^{2}}=25 \\  & {{x}^{2}}+{{y}^{2}}-12x+11=0 \\ \end{align}$

So, the equation of the circle can be ${{x}^{2}}+{{y}^{2}}+4x-21=0 $ or ${{x}^{2}}+{{y}^{2}}-12x+11=0 $

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through (0,0).

$\begin{align}  & {{(0-h)}^{2}}+{{(0-k)}^{2}}={{r}^{2}} \\  & \Rightarrow {{h}^{2}}+{{k}^{2}}={{r}^{2}} \\ \end{align}$

The equation of the circle now becomes ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{h}^{2}}+{{k}^{2}}$

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points $(a,0)$ and $(0,b)$. Therefore,

${{(a-h)}^{2}}+{{(0-k)}^{2}}={{h}^{2}}+{{k}^{2}}$ …….(1)

${{(0-h)}^{2}}+{{(b-k)}^{2}}={{h}^{2}}+{{k}^{2}}$ …….(2)

From equation (1), we’ll get

$\begin{align}  & {{a}^{2}}-2ah+{{h}^{2}}+{{k}^{2}}={{h}^{2}}+{{k}^{2}} \\  & \Rightarrow {{a}^{2}}-2ah=0 \\  & \Rightarrow a(a-2h)=0 \\ \end{align}$

$\Rightarrow a=0$ or $(a-2h)=0$

However, $a\ne 0;$ hence, $(a-2h)=0\Rightarrow h=\dfrac{a}{2}$

From equation (2), we’ll get

$\begin{align}  & {{h}^{2}}+{{b}^{2}}-2bk+{{k}^{2}}={{h}^{2}}+{{k}^{2}} \\  & \Rightarrow {{b}^{2}}-2bk=0 \\  & \Rightarrow b(b-2k)=0 \\ \end{align}$

$\Rightarrow b=0$ or $(b-2k)=0$

However, $b\ne 0;$ hence, $(b-2k)=0\Rightarrow k=\dfrac{b}{2}$

Thus, the equation of the required circle is

${{\left( x-\dfrac{a}{2} \right)}^{2}}+{{\left( y-\dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( \dfrac{b}{2} \right)}^{2}}$

${{\left( \dfrac{2x-a}{2} \right)}^{2}}+{{\left( \dfrac{2y-b}{2} \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{4}$

$\begin{align}  & \Rightarrow 4{{x}^{2}}-4ax+{{a}^{2}}+4{{y}^{2}}-4by+{{b}^{2}}={{a}^{2}}+{{b}^{2}} \\  & \Rightarrow 4{{x}^{2}}+4{{y}^{2}}-4ax-4by=0 \\  & \Rightarrow {{x}^{2}}+{{y}^{2}}-ax-by=0 \\ \end{align}$

Ans: The centre of the circle is given as (h,k)=(2,2)

Since the circle passes through point (4,5), the radius (r) of the circle is the distance between the points (2,2) and (4,5).

$\begin{align}  & \therefore r=\sqrt{{{(2-4)}^{2}}+{{(2-5)}^{2}}} \\  & =\sqrt{{{(-2)}^{2}}+{{(-3)}^{2}}} \\  & =\sqrt{4+9} \\  & =\sqrt{13} \\ \end{align}$

Thus, the equation of the circle is 

${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

$\begin{align}  & {{(x-2)}^{2}}+{{(y-2)}^{2}}={{\left( \sqrt{13} \right)}^{2}} \\  & {{x}^{2}}-4x+4+{{y}^{2}}-4y+4=13 \\  & {{x}^{2}}+{{y}^{2}}-4x-4y-5=0 \\ \end{align}$

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}=25$.

${{x}^{2}}+{{y}^{2}}=25$

$\Rightarrow {{(x-0)}^{2}}+{{(y-0)}^{2}}={{5}^{2}},$ which is of the form of ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h = 0, k = 0 \; and\; r=5$

$\therefore $ Centre =(0,0) and radius =5

Distance between point (-2.5, 3.5) and centre (0,0)

$=\sqrt{{{(-2.5-0)}^{2}}+{{(3.5-0)}^{2}}}$

$\begin{align}  & =\sqrt{6.25+12.25} \\  & =\sqrt{18.25} \\ \end{align}$

$=4.3$ (approx.) $<5$

Since the distance between point (-2.5, 3.5) and centre (0,0) of the circle is less than the radius of the circle, point (-2.5, 3.5) lies inside the circle.